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  #46   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

chung wrote in message ervers.com...
Svante wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


Several reasons:

1. 40 dB down is 40 dB down, whether you're talking about voltage or
power, assuming constant load impedance. If the 2nd harmonic is 40 dB
down, it means the voltage ratio is 1%, and the ratio of delivered power
is 0.01%. A dB in voltage is a dB in power!


Yes, I am aware of this, and perhaps that is why I like expressing
distorsion in dB rather than as a percentage. "A dB is a dB". Or is
it? Hmm, se below!

2. Audio amplifiers are voltage devices. The actual power delivered to
the load depends on the load impedance. For example, let's say an
amplifer has 1% 2nd harmonic distortion in voltage. How much power is
delivered to the load at that 2nd harmonic frequency? The answer depends
on the load impedance at that frequency. It is not unusual for a
speaker's impedance to change substantially over an octave. So in this
case, the power ratio may not be 0.01%.


That is the best explanation so far, and the only one that doesn't
rely on a convention, but rather a physical fact (a frequency
dependent load resistance). However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.

3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load? I mean, the fundaments of dB
assumes that we measure a power ratio. The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance. It simply assumes that the load
resistenace is constant.

Hmmm... Interesting.
  #47   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

chung wrote in message ervers.com...
Svante wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


Several reasons:

1. 40 dB down is 40 dB down, whether you're talking about voltage or
power, assuming constant load impedance. If the 2nd harmonic is 40 dB
down, it means the voltage ratio is 1%, and the ratio of delivered power
is 0.01%. A dB in voltage is a dB in power!


Yes, I am aware of this, and perhaps that is why I like expressing
distorsion in dB rather than as a percentage. "A dB is a dB". Or is
it? Hmm, se below!

2. Audio amplifiers are voltage devices. The actual power delivered to
the load depends on the load impedance. For example, let's say an
amplifer has 1% 2nd harmonic distortion in voltage. How much power is
delivered to the load at that 2nd harmonic frequency? The answer depends
on the load impedance at that frequency. It is not unusual for a
speaker's impedance to change substantially over an octave. So in this
case, the power ratio may not be 0.01%.


That is the best explanation so far, and the only one that doesn't
rely on a convention, but rather a physical fact (a frequency
dependent load resistance). However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.

3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load? I mean, the fundaments of dB
assumes that we measure a power ratio. The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance. It simply assumes that the load
resistenace is constant.

Hmmm... Interesting.
  #48   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

chung wrote in message ervers.com...
Svante wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


Several reasons:

1. 40 dB down is 40 dB down, whether you're talking about voltage or
power, assuming constant load impedance. If the 2nd harmonic is 40 dB
down, it means the voltage ratio is 1%, and the ratio of delivered power
is 0.01%. A dB in voltage is a dB in power!


Yes, I am aware of this, and perhaps that is why I like expressing
distorsion in dB rather than as a percentage. "A dB is a dB". Or is
it? Hmm, se below!

2. Audio amplifiers are voltage devices. The actual power delivered to
the load depends on the load impedance. For example, let's say an
amplifer has 1% 2nd harmonic distortion in voltage. How much power is
delivered to the load at that 2nd harmonic frequency? The answer depends
on the load impedance at that frequency. It is not unusual for a
speaker's impedance to change substantially over an octave. So in this
case, the power ratio may not be 0.01%.


That is the best explanation so far, and the only one that doesn't
rely on a convention, but rather a physical fact (a frequency
dependent load resistance). However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.

3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load? I mean, the fundaments of dB
assumes that we measure a power ratio. The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance. It simply assumes that the load
resistenace is constant.

Hmmm... Interesting.
  #49   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

chung wrote in message ervers.com...
Svante wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


Several reasons:

1. 40 dB down is 40 dB down, whether you're talking about voltage or
power, assuming constant load impedance. If the 2nd harmonic is 40 dB
down, it means the voltage ratio is 1%, and the ratio of delivered power
is 0.01%. A dB in voltage is a dB in power!


Yes, I am aware of this, and perhaps that is why I like expressing
distorsion in dB rather than as a percentage. "A dB is a dB". Or is
it? Hmm, se below!

2. Audio amplifiers are voltage devices. The actual power delivered to
the load depends on the load impedance. For example, let's say an
amplifer has 1% 2nd harmonic distortion in voltage. How much power is
delivered to the load at that 2nd harmonic frequency? The answer depends
on the load impedance at that frequency. It is not unusual for a
speaker's impedance to change substantially over an octave. So in this
case, the power ratio may not be 0.01%.


That is the best explanation so far, and the only one that doesn't
rely on a convention, but rather a physical fact (a frequency
dependent load resistance). However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.

3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load? I mean, the fundaments of dB
assumes that we measure a power ratio. The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance. It simply assumes that the load
resistenace is constant.

Hmmm... Interesting.
  #54   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Dick Pierce) wrote in message . com...
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.


Precisely. ALL of the THD meters in existance are essentially
voltmeters with filters. Going from the ancient and venerable
General Radio 1936 or Hewlett-Packard 330, through the HP 334,
through the Sound Technology 1700, Amber, Marconi and so on,
they are ALL voltmeters that have a narrow-band notch filter
in them.


Hmm, just curious, did they have circuitry to measure RMS voltage,
rather than average voltage? I guess so, because that would be crucial
to end up with a correct result. The point here is that the only
reasonable way to add up harmonics is to add their powers. That is why
the *RMS* voltage of the (sum of all) harmonics would have to be
measured.

And, if you describe the ratio of the original to its distortion
products in dB rather than in percentage, what do you care?


Exactly! That is (part of) why I like dBs!

A THD of 1% means the amplitudes differ by a factor of 100:1,
their powers differ by 10000:1, but it's 40 dB in both cases.
That's the entire point: A RATIO between two signals is EXACTLY
the same whether you describe the RATIO of the voltage, the RATIO
of the power or the LOGS of those ratios (assuming the impedance
is the same in all cases, which is a prefectly reasonable assumption).


Yes I am aware of this, but one cannot point it out too often.

There is no more or less logic to doing it one way or the other,
they are exactly equivalent.


Yes, and the risk for misunderstanding speaks for measuring distortion
in dBs rather than in %, IMO.

Isn't it surprising that no marketers have found the opportunity to
say that eg a loudspeaker has a distorsion of loudspeaker is 0.01%
(power ratio) rather than 1% (voltage ratio). Both could in a sense be
correct, and correspond to -40 dB. Oh, I better shut up.
  #55   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Dick Pierce) wrote in message . com...
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.


Precisely. ALL of the THD meters in existance are essentially
voltmeters with filters. Going from the ancient and venerable
General Radio 1936 or Hewlett-Packard 330, through the HP 334,
through the Sound Technology 1700, Amber, Marconi and so on,
they are ALL voltmeters that have a narrow-band notch filter
in them.


Hmm, just curious, did they have circuitry to measure RMS voltage,
rather than average voltage? I guess so, because that would be crucial
to end up with a correct result. The point here is that the only
reasonable way to add up harmonics is to add their powers. That is why
the *RMS* voltage of the (sum of all) harmonics would have to be
measured.

And, if you describe the ratio of the original to its distortion
products in dB rather than in percentage, what do you care?


Exactly! That is (part of) why I like dBs!

A THD of 1% means the amplitudes differ by a factor of 100:1,
their powers differ by 10000:1, but it's 40 dB in both cases.
That's the entire point: A RATIO between two signals is EXACTLY
the same whether you describe the RATIO of the voltage, the RATIO
of the power or the LOGS of those ratios (assuming the impedance
is the same in all cases, which is a prefectly reasonable assumption).


Yes I am aware of this, but one cannot point it out too often.

There is no more or less logic to doing it one way or the other,
they are exactly equivalent.


Yes, and the risk for misunderstanding speaks for measuring distortion
in dBs rather than in %, IMO.

Isn't it surprising that no marketers have found the opportunity to
say that eg a loudspeaker has a distorsion of loudspeaker is 0.01%
(power ratio) rather than 1% (voltage ratio). Both could in a sense be
correct, and correspond to -40 dB. Oh, I better shut up.


  #56   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Dick Pierce) wrote in message . com...
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.


Precisely. ALL of the THD meters in existance are essentially
voltmeters with filters. Going from the ancient and venerable
General Radio 1936 or Hewlett-Packard 330, through the HP 334,
through the Sound Technology 1700, Amber, Marconi and so on,
they are ALL voltmeters that have a narrow-band notch filter
in them.


Hmm, just curious, did they have circuitry to measure RMS voltage,
rather than average voltage? I guess so, because that would be crucial
to end up with a correct result. The point here is that the only
reasonable way to add up harmonics is to add their powers. That is why
the *RMS* voltage of the (sum of all) harmonics would have to be
measured.

And, if you describe the ratio of the original to its distortion
products in dB rather than in percentage, what do you care?


Exactly! That is (part of) why I like dBs!

A THD of 1% means the amplitudes differ by a factor of 100:1,
their powers differ by 10000:1, but it's 40 dB in both cases.
That's the entire point: A RATIO between two signals is EXACTLY
the same whether you describe the RATIO of the voltage, the RATIO
of the power or the LOGS of those ratios (assuming the impedance
is the same in all cases, which is a prefectly reasonable assumption).


Yes I am aware of this, but one cannot point it out too often.

There is no more or less logic to doing it one way or the other,
they are exactly equivalent.


Yes, and the risk for misunderstanding speaks for measuring distortion
in dBs rather than in %, IMO.

Isn't it surprising that no marketers have found the opportunity to
say that eg a loudspeaker has a distorsion of loudspeaker is 0.01%
(power ratio) rather than 1% (voltage ratio). Both could in a sense be
correct, and correspond to -40 dB. Oh, I better shut up.
  #57   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Dick Pierce) wrote in message . com...
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.


Precisely. ALL of the THD meters in existance are essentially
voltmeters with filters. Going from the ancient and venerable
General Radio 1936 or Hewlett-Packard 330, through the HP 334,
through the Sound Technology 1700, Amber, Marconi and so on,
they are ALL voltmeters that have a narrow-band notch filter
in them.


Hmm, just curious, did they have circuitry to measure RMS voltage,
rather than average voltage? I guess so, because that would be crucial
to end up with a correct result. The point here is that the only
reasonable way to add up harmonics is to add their powers. That is why
the *RMS* voltage of the (sum of all) harmonics would have to be
measured.

And, if you describe the ratio of the original to its distortion
products in dB rather than in percentage, what do you care?


Exactly! That is (part of) why I like dBs!

A THD of 1% means the amplitudes differ by a factor of 100:1,
their powers differ by 10000:1, but it's 40 dB in both cases.
That's the entire point: A RATIO between two signals is EXACTLY
the same whether you describe the RATIO of the voltage, the RATIO
of the power or the LOGS of those ratios (assuming the impedance
is the same in all cases, which is a prefectly reasonable assumption).


Yes I am aware of this, but one cannot point it out too often.

There is no more or less logic to doing it one way or the other,
they are exactly equivalent.


Yes, and the risk for misunderstanding speaks for measuring distortion
in dBs rather than in %, IMO.

Isn't it surprising that no marketers have found the opportunity to
say that eg a loudspeaker has a distorsion of loudspeaker is 0.01%
(power ratio) rather than 1% (voltage ratio). Both could in a sense be
correct, and correspond to -40 dB. Oh, I better shut up.
  #60   Report Post  
Stewart Pinkerton
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.


Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................

But yes, I still *think* in feet, not metres................
--

Stewart Pinkerton | Music is Art - Audio is Engineering


  #61   Report Post  
Stewart Pinkerton
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.


Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................

But yes, I still *think* in feet, not metres................
--

Stewart Pinkerton | Music is Art - Audio is Engineering
  #64   Report Post  
Stewart Pinkerton
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.


Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................

But yes, I still *think* in feet, not metres................
--

Stewart Pinkerton | Music is Art - Audio is Engineering
  #65   Report Post  
Stewart Pinkerton
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.


Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................

But yes, I still *think* in feet, not metres................
--

Stewart Pinkerton | Music is Art - Audio is Engineering


  #66   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.


Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


---
To obtain a standard of length a quadrant of the earth (one-fourth of a
circumference) was surveyed from Dunkirk to Barcelona along the meridian
that passes through Paris. The distance from the pole to the equator was
divided into ten million parts to constitute the meter (spelled metre in
some countries).

Further subdivisions in the length of the meter, by orders of magnitude,
into more convenient-to-use units for some applications led to the
naming of the decimeter (one-tenth of a meter), the centimeter (one
one-hundredth of a meter), the millimeter (one one-thousandth of a
meter) and so on. Note that by defining the unit of length the
definition of the unit of volume followed automatically.

Note also the curious coincidence(?) of units in the metric system being
divisible everywhere by ten and the fact that we have ten digits on our
hands.

Now, since water is/was ubiquitous on the surface of the earth and,
presumably, weighed the same everywhere, it was decided that a certain
volume of water (the 'cubic centimeter', a cube one centimeter on an
edge) would become the standard of weight and was called the 'gramme'.

The prefix 'kilo', indicating that a multiplication of the quantity
following it by 1000 is required, means "1000 grams" when appended with
'gram'. Hence, kilo+gram = kilogram = 1000 * 1 gram = 1000 grams.

--
John Fields
  #67   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.


Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


---
To obtain a standard of length a quadrant of the earth (one-fourth of a
circumference) was surveyed from Dunkirk to Barcelona along the meridian
that passes through Paris. The distance from the pole to the equator was
divided into ten million parts to constitute the meter (spelled metre in
some countries).

Further subdivisions in the length of the meter, by orders of magnitude,
into more convenient-to-use units for some applications led to the
naming of the decimeter (one-tenth of a meter), the centimeter (one
one-hundredth of a meter), the millimeter (one one-thousandth of a
meter) and so on. Note that by defining the unit of length the
definition of the unit of volume followed automatically.

Note also the curious coincidence(?) of units in the metric system being
divisible everywhere by ten and the fact that we have ten digits on our
hands.

Now, since water is/was ubiquitous on the surface of the earth and,
presumably, weighed the same everywhere, it was decided that a certain
volume of water (the 'cubic centimeter', a cube one centimeter on an
edge) would become the standard of weight and was called the 'gramme'.

The prefix 'kilo', indicating that a multiplication of the quantity
following it by 1000 is required, means "1000 grams" when appended with
'gram'. Hence, kilo+gram = kilogram = 1000 * 1 gram = 1000 grams.

--
John Fields
  #68   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.


Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


---
To obtain a standard of length a quadrant of the earth (one-fourth of a
circumference) was surveyed from Dunkirk to Barcelona along the meridian
that passes through Paris. The distance from the pole to the equator was
divided into ten million parts to constitute the meter (spelled metre in
some countries).

Further subdivisions in the length of the meter, by orders of magnitude,
into more convenient-to-use units for some applications led to the
naming of the decimeter (one-tenth of a meter), the centimeter (one
one-hundredth of a meter), the millimeter (one one-thousandth of a
meter) and so on. Note that by defining the unit of length the
definition of the unit of volume followed automatically.

Note also the curious coincidence(?) of units in the metric system being
divisible everywhere by ten and the fact that we have ten digits on our
hands.

Now, since water is/was ubiquitous on the surface of the earth and,
presumably, weighed the same everywhere, it was decided that a certain
volume of water (the 'cubic centimeter', a cube one centimeter on an
edge) would become the standard of weight and was called the 'gramme'.

The prefix 'kilo', indicating that a multiplication of the quantity
following it by 1000 is required, means "1000 grams" when appended with
'gram'. Hence, kilo+gram = kilogram = 1000 * 1 gram = 1000 grams.

--
John Fields
  #69   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.


Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


---
To obtain a standard of length a quadrant of the earth (one-fourth of a
circumference) was surveyed from Dunkirk to Barcelona along the meridian
that passes through Paris. The distance from the pole to the equator was
divided into ten million parts to constitute the meter (spelled metre in
some countries).

Further subdivisions in the length of the meter, by orders of magnitude,
into more convenient-to-use units for some applications led to the
naming of the decimeter (one-tenth of a meter), the centimeter (one
one-hundredth of a meter), the millimeter (one one-thousandth of a
meter) and so on. Note that by defining the unit of length the
definition of the unit of volume followed automatically.

Note also the curious coincidence(?) of units in the metric system being
divisible everywhere by ten and the fact that we have ten digits on our
hands.

Now, since water is/was ubiquitous on the surface of the earth and,
presumably, weighed the same everywhere, it was decided that a certain
volume of water (the 'cubic centimeter', a cube one centimeter on an
edge) would become the standard of weight and was called the 'gramme'.

The prefix 'kilo', indicating that a multiplication of the quantity
following it by 1000 is required, means "1000 grams" when appended with
'gram'. Hence, kilo+gram = kilogram = 1000 * 1 gram = 1000 grams.

--
John Fields
  #70   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:02:57 -0800,
(Svante)
wrote:

However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.


The dB was originally a measure of sound pressure level, and the
logarithmic scale is used simply becuause our ears respond to sound in
a logarithmic fashion.


Nope. Think a bit. "deci" is a tenth. Why would mister Bell have
defined a Bel as TWO times the logarithm of the ratio between two
voltages/pressures/currents?
The original definition of Bel is simply (ONE time) the tenth
logarithm of a power ratio. Then, if we want to end up with the same
number measuring voltages, we will have to take the logarithm of the
SQUARE of the voltage ratio, which is the same as TWO times the
logarithm of the voltage ratio. Or:

Bel = log(p/pref) = log((u/uref)^2 = 2 * log(u/uref)
or:
deciBel= 10 * log(p/pref) = 10 * log((u/uref)^2 = 20 * log(u/uref)

....given that pref and uref corresponds to the same power in the
resistance in question.

The "2 (* log...)" is a simple by-product of the square in P=U^2/R

So I guess it is safe to assume that the fundament of the deciBel
rests on a power ratio.


3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load?


Not at all, since a voltage ratio of 2:1 is approximately 6dB,
regardless of current flow. This is why voltage, not power, is used
as a standard measure of speaker sensitivity, since it is independent
of the load impedance.


Not if the decibel indicates the power ratio.


I mean, the fundaments of dB
assumes that we measure a power ratio.


No, it doesn't. It is simply a useful logarithmic ratio.


Indeed, the Bel is a useful logarithm of a power ratio.


The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance. It simply assumes that the load
resistenace is constant.


It assumes no such thing.


If you get back to the fundaments, it does. But I know what you mean,
I am just trying to point out that on the way to the voltage ratio,
somewhere, the resistance has been ignored.


  #71   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:02:57 -0800,
(Svante)
wrote:

However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.


The dB was originally a measure of sound pressure level, and the
logarithmic scale is used simply becuause our ears respond to sound in
a logarithmic fashion.


Nope. Think a bit. "deci" is a tenth. Why would mister Bell have
defined a Bel as TWO times the logarithm of the ratio between two
voltages/pressures/currents?
The original definition of Bel is simply (ONE time) the tenth
logarithm of a power ratio. Then, if we want to end up with the same
number measuring voltages, we will have to take the logarithm of the
SQUARE of the voltage ratio, which is the same as TWO times the
logarithm of the voltage ratio. Or:

Bel = log(p/pref) = log((u/uref)^2 = 2 * log(u/uref)
or:
deciBel= 10 * log(p/pref) = 10 * log((u/uref)^2 = 20 * log(u/uref)

....given that pref and uref corresponds to the same power in the
resistance in question.

The "2 (* log...)" is a simple by-product of the square in P=U^2/R

So I guess it is safe to assume that the fundament of the deciBel
rests on a power ratio.


3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load?


Not at all, since a voltage ratio of 2:1 is approximately 6dB,
regardless of current flow. This is why voltage, not power, is used
as a standard measure of speaker sensitivity, since it is independent
of the load impedance.


Not if the decibel indicates the power ratio.


I mean, the fundaments of dB
assumes that we measure a power ratio.


No, it doesn't. It is simply a useful logarithmic ratio.


Indeed, the Bel is a useful logarithm of a power ratio.


The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance. It simply assumes that the load
resistenace is constant.


It assumes no such thing.


If you get back to the fundaments, it does. But I know what you mean,
I am just trying to point out that on the way to the voltage ratio,
somewhere, the resistance has been ignored.
  #72   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:02:57 -0800,
(Svante)
wrote:

However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.


The dB was originally a measure of sound pressure level, and the
logarithmic scale is used simply becuause our ears respond to sound in
a logarithmic fashion.


Nope. Think a bit. "deci" is a tenth. Why would mister Bell have
defined a Bel as TWO times the logarithm of the ratio between two
voltages/pressures/currents?
The original definition of Bel is simply (ONE time) the tenth
logarithm of a power ratio. Then, if we want to end up with the same
number measuring voltages, we will have to take the logarithm of the
SQUARE of the voltage ratio, which is the same as TWO times the
logarithm of the voltage ratio. Or:

Bel = log(p/pref) = log((u/uref)^2 = 2 * log(u/uref)
or:
deciBel= 10 * log(p/pref) = 10 * log((u/uref)^2 = 20 * log(u/uref)

....given that pref and uref corresponds to the same power in the
resistance in question.

The "2 (* log...)" is a simple by-product of the square in P=U^2/R

So I guess it is safe to assume that the fundament of the deciBel
rests on a power ratio.


3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load?


Not at all, since a voltage ratio of 2:1 is approximately 6dB,
regardless of current flow. This is why voltage, not power, is used
as a standard measure of speaker sensitivity, since it is independent
of the load impedance.


Not if the decibel indicates the power ratio.


I mean, the fundaments of dB
assumes that we measure a power ratio.


No, it doesn't. It is simply a useful logarithmic ratio.


Indeed, the Bel is a useful logarithm of a power ratio.


The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance. It simply assumes that the load
resistenace is constant.


It assumes no such thing.


If you get back to the fundaments, it does. But I know what you mean,
I am just trying to point out that on the way to the voltage ratio,
somewhere, the resistance has been ignored.
  #73   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:02:57 -0800,
(Svante)
wrote:

However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.


The dB was originally a measure of sound pressure level, and the
logarithmic scale is used simply becuause our ears respond to sound in
a logarithmic fashion.


Nope. Think a bit. "deci" is a tenth. Why would mister Bell have
defined a Bel as TWO times the logarithm of the ratio between two
voltages/pressures/currents?
The original definition of Bel is simply (ONE time) the tenth
logarithm of a power ratio. Then, if we want to end up with the same
number measuring voltages, we will have to take the logarithm of the
SQUARE of the voltage ratio, which is the same as TWO times the
logarithm of the voltage ratio. Or:

Bel = log(p/pref) = log((u/uref)^2 = 2 * log(u/uref)
or:
deciBel= 10 * log(p/pref) = 10 * log((u/uref)^2 = 20 * log(u/uref)

....given that pref and uref corresponds to the same power in the
resistance in question.

The "2 (* log...)" is a simple by-product of the square in P=U^2/R

So I guess it is safe to assume that the fundament of the deciBel
rests on a power ratio.


3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load?


Not at all, since a voltage ratio of 2:1 is approximately 6dB,
regardless of current flow. This is why voltage, not power, is used
as a standard measure of speaker sensitivity, since it is independent
of the load impedance.


Not if the decibel indicates the power ratio.


I mean, the fundaments of dB
assumes that we measure a power ratio.


No, it doesn't. It is simply a useful logarithmic ratio.


Indeed, the Bel is a useful logarithm of a power ratio.


The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance. It simply assumes that the load
resistenace is constant.


It assumes no such thing.


If you get back to the fundaments, it does. But I know what you mean,
I am just trying to point out that on the way to the voltage ratio,
somewhere, the resistance has been ignored.
  #74   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:06:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................

But yes, I still *think* in feet, not metres................


Should I say that I'm a Swede, if you wonder. I was raised with the
metric system, and appreciate it a lot and cannot understand how feet
could possibly be more intuitive than metres. Heck, even our mile is
metric (10 000 metres).
I can't help quoting my mechanics teacher who was a member of the
swedish standards committe: "England is going metric, inch by inch..."
  #75   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:06:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................

But yes, I still *think* in feet, not metres................


Should I say that I'm a Swede, if you wonder. I was raised with the
metric system, and appreciate it a lot and cannot understand how feet
could possibly be more intuitive than metres. Heck, even our mile is
metric (10 000 metres).
I can't help quoting my mechanics teacher who was a member of the
swedish standards committe: "England is going metric, inch by inch..."


  #76   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:06:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................

But yes, I still *think* in feet, not metres................


Should I say that I'm a Swede, if you wonder. I was raised with the
metric system, and appreciate it a lot and cannot understand how feet
could possibly be more intuitive than metres. Heck, even our mile is
metric (10 000 metres).
I can't help quoting my mechanics teacher who was a member of the
swedish standards committe: "England is going metric, inch by inch..."
  #77   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:06:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................

But yes, I still *think* in feet, not metres................


Should I say that I'm a Swede, if you wonder. I was raised with the
metric system, and appreciate it a lot and cannot understand how feet
could possibly be more intuitive than metres. Heck, even our mile is
metric (10 000 metres).
I can't help quoting my mechanics teacher who was a member of the
swedish standards committe: "England is going metric, inch by inch..."
  #78   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 16 Jan 2004 19:04:44 -0800, (Dick Pierce)
wrote:

John Fields wrote in message . ..
On 16 Jan 2004 14:32:02 -0800,
(Dick Pierce)
wrote:

John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:


And, if you describe the ratio of the original to its distortion
products in dB rather than in percentage, what do you care?
A THD of 1% means the amplitudes differ by a factor of 100:1,
their powers differ by 10000:1, but it's 40 dB in both cases.
That's the entire point: A RATIO between two signals is EXACTLY
the same whether you describe the RATIO of the voltage, the RATIO
of the power or the LOGS of those ratios (assuming the impedance
is the same in all cases, which is a prefectly reasonable assumption).


---
Except that decibels describing the ratio of one power to another

P1
is dB = 10 log ---- , while for voltages or currents it's _20_ times
P2
the log of the ratio.


But WHY? It's simply because power is proportional to the square of
the voltage, thus since:

dB = 10 log P1/Pref

and since

P prop V^2

and thus:

dB = 10 log V1^2/Vref^2

or

dB = 10 log (V1/Vref)^2

and

dB = 2 * 10 log (V1/Vref)

thus

dB = 20 log V1/Vref

QED.

So, it stands that 40 dB is 40 dB, whether we started with the
ratio of two voltages, or the ratio of the equivalent powers
of those two voltages.


---
Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

if V = 2Vref, then dB(V) = 20 log (2/1) ~ 6dB.


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

--
John Fields
  #79   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 16 Jan 2004 19:04:44 -0800, (Dick Pierce)
wrote:

John Fields wrote in message . ..
On 16 Jan 2004 14:32:02 -0800,
(Dick Pierce)
wrote:

John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:


And, if you describe the ratio of the original to its distortion
products in dB rather than in percentage, what do you care?
A THD of 1% means the amplitudes differ by a factor of 100:1,
their powers differ by 10000:1, but it's 40 dB in both cases.
That's the entire point: A RATIO between two signals is EXACTLY
the same whether you describe the RATIO of the voltage, the RATIO
of the power or the LOGS of those ratios (assuming the impedance
is the same in all cases, which is a prefectly reasonable assumption).


---
Except that decibels describing the ratio of one power to another

P1
is dB = 10 log ---- , while for voltages or currents it's _20_ times
P2
the log of the ratio.


But WHY? It's simply because power is proportional to the square of
the voltage, thus since:

dB = 10 log P1/Pref

and since

P prop V^2

and thus:

dB = 10 log V1^2/Vref^2

or

dB = 10 log (V1/Vref)^2

and

dB = 2 * 10 log (V1/Vref)

thus

dB = 20 log V1/Vref

QED.

So, it stands that 40 dB is 40 dB, whether we started with the
ratio of two voltages, or the ratio of the equivalent powers
of those two voltages.


---
Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

if V = 2Vref, then dB(V) = 20 log (2/1) ~ 6dB.


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

--
John Fields
  #80   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 16 Jan 2004 19:04:44 -0800, (Dick Pierce)
wrote:

John Fields wrote in message . ..
On 16 Jan 2004 14:32:02 -0800,
(Dick Pierce)
wrote:

John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:


And, if you describe the ratio of the original to its distortion
products in dB rather than in percentage, what do you care?
A THD of 1% means the amplitudes differ by a factor of 100:1,
their powers differ by 10000:1, but it's 40 dB in both cases.
That's the entire point: A RATIO between two signals is EXACTLY
the same whether you describe the RATIO of the voltage, the RATIO
of the power or the LOGS of those ratios (assuming the impedance
is the same in all cases, which is a prefectly reasonable assumption).


---
Except that decibels describing the ratio of one power to another

P1
is dB = 10 log ---- , while for voltages or currents it's _20_ times
P2
the log of the ratio.


But WHY? It's simply because power is proportional to the square of
the voltage, thus since:

dB = 10 log P1/Pref

and since

P prop V^2

and thus:

dB = 10 log V1^2/Vref^2

or

dB = 10 log (V1/Vref)^2

and

dB = 2 * 10 log (V1/Vref)

thus

dB = 20 log V1/Vref

QED.

So, it stands that 40 dB is 40 dB, whether we started with the
ratio of two voltages, or the ratio of the equivalent powers
of those two voltages.


---
Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

if V = 2Vref, then dB(V) = 20 log (2/1) ~ 6dB.


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

--
John Fields


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